Simplify and expand the following expression: $ \dfrac{1}{k - 9}- \dfrac{3}{4k - 24}- \dfrac{3}{k^2 - 15k + 54} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the second term: $ \dfrac{3}{4k - 24} = \dfrac{3}{4(k - 6)}$ We can factor the quadratic in the third term: $ \dfrac{3}{k^2 - 15k + 54} = \dfrac{3}{(k - 9)(k - 6)}$ Now we have: $ \dfrac{1}{k - 9}- \dfrac{3}{4(k - 6)}- \dfrac{3}{(k - 9)(k - 6)} $ The least common multiple of the denominators is: $ (k - 9)(k - 6)$ In order to get the first term over $(k - 9)(k - 6)$ , multiply by $\dfrac{4(k - 6)}{4(k - 6)}$ $ \dfrac{1}{k - 9} \times \dfrac{4(k - 6)}{4(k - 6)} = \dfrac{4(k - 6)}{(k - 9)(k - 6)} $ In order to get the second term over $(k - 9)(k - 6)$ , multiply by $\dfrac{k - 9}{k - 9}$ $ \dfrac{3}{4(k - 6)} \times \dfrac{k - 9}{k - 9} = \dfrac{3(k - 9)}{(k - 9)(k - 6)} $ In order to get the third term over $(k - 9)(k - 6)$ , multiply by $\dfrac{4}{4}$ $ \dfrac{3}{(k - 9)(k - 6)} \times \dfrac{4}{4} = \dfrac{12}{(k - 9)(k - 6)} $ Now we have: $ \dfrac{4(k - 6)}{(k - 9)(k - 6)} - \dfrac{3(k - 9)}{(k - 9)(k - 6)} - \dfrac{12}{(k - 9)(k - 6)} $ $ = \dfrac{ 4(k - 6) - 3(k - 9) - 12} {(k - 9)(k - 6)} $ Expand: $ = \dfrac{4k - 24 - 3k + 27 - 12}{4k^2 - 60k + 216} $ $ = \dfrac{k - 9}{4k^2 - 60k + 216}$